MCQ
A glass wind screen whose inclination with the vertical can be changed is mounted on a car. The car moves horizontally with a speed of $2\,\,m/s$. At what angle $\alpha$ with the vertical should the wind screen be placed so that the rain drops falling vertically downwards with velocity $6\,\, m/s$ strike the wind screen perpendicularly.
  • $tan^{-1}(3)$
  • B
    $tan^{-1}(1/3)$
  • C
    $cos^{^{-1}}(3)$
  • D
    $sin^{-1}(1/3)$

Answer

Correct option: A.
$tan^{-1}(3)$
a
Let vertical direction limit vector be $j,$ horizontal

$\mathrm{V}_{\mathrm{car}}=2 \hat{\mathrm{i}}$

$V_{\text {drops }}=6 \hat{j}$

$\mathrm{V}_{\text {drops }} \mathrm{W} .$ $r.t$ car $=\mathrm{V}_{\text {drops }}-\mathrm{V}_{\text {car }}$

$=-2 \hat{i}+6 \hat{j}$

$\cos \theta_{1}=\frac{\hat{j} \cdot(-2 \hat{i}+6 \hat{j})}{|\hat{j}| \cdot|-2 \hat{i}+6 \hat{j}|}$

$=\frac{6}{\sqrt{2^{2}+6^{2}} \cdot 1}=\frac{6}{\sqrt{40}}$

$\Rightarrow \cos \theta_{1}=\frac{3}{\sqrt{10}}$

$\Rightarrow \sin \theta_{2}=\frac{3}{\sqrt{10}}$

$\Rightarrow \tan \theta_{2}=3$

$\Rightarrow \theta_{2}=\tan ^{-1} 3$

Hence wind stream is inclined at tan $^{-1} 3^{\circ}$ with vertical.

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